Question

Cho N = \(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+ ............. +\(\frac{1}{1000\cdot1001}\) so sánh N với M = \(\frac{2019}{2020}\)

Giup với mình đang cần gấp!

Answer 1

Ta có : \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1000.1001}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1001-1000}{1000.1001}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)

\(=1-\frac{1}{1001}=\frac{1000}{1001}\)

Ta thấy : \(1001< 2020\Rightarrow\frac{1}{1001}>\frac{1}{2020}\)

\(\Rightarrow-\frac{1}{1001}< -\frac{1}{2020}\)

\(\Rightarrow1-\frac{1}{1001}< 1-\frac{1}{2020}\Rightarrow\frac{1000}{1001}< \frac{2019}{2020}\)

Hay : \(N< M\)

Answer 2

Lộn đề M = \(\frac{20192019}{20202020}\)NHA

Answer 3

\(N=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1000}-\frac{1}{1001}\) 

\(=\frac{1}{1}-\frac{1}{1001}\) 

\(=\frac{1000}{1001}\) 

\(\frac{1000}{1001}=\frac{1000\cdot2020}{1001\cdot2020}\) 

\(\frac{2019}{2020}=\frac{2019\cdot1001}{1001\cdot2020}=\frac{2019\cdot1001-2019+1001+1018}{1001\cdot2020}=\frac{\left(1001\cdot2020\right)+1018}{1001\cdot2020}\)  

Vậy N < M